document.write( "Question 99385This question is from textbook
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document.write( ": Solve the equation by completing the square, giving your solutions exactly in surd form\r
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document.write( "2x^2 - 10x + 5 = 0\r
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document.write( "i got so far but seem to be mucking up somewhere\r
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document.write( "2(x^2 - 5x + 5/2) = 0\r
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document.write( "(x^2 - 5/2)^2 = x^2 - 5x + 25/4\r
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document.write( "but now i'm stuck please could you help\r
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document.write( "regards, nat x \n" );
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Algebra.Com's Answer #72386 by scott8148(6628)![]() ![]() You can put this solution on YOUR website! first, using multiplication or division, set the coefficient of the squared term to 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "2x^2-10x+5=0 ... dividing by 2 gives x^2-5x+(5/2)=0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "next, using addition or subtraction, set the constant term to 0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x^2-5x+(5/2)=0 ... subtracting 5/2 gives x^2-5x=-5/2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "next, take 1/2 of the coefficient of the first order term, square it, and add (to BOTH sides)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(-5/2)^2=25/4 ... x^2-5x+(25/4)=15/4\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "take the square root (positive AND negative) ... x-(5/2)=(sqrt(15))/2 and x-(5/2)=-(sqrt(15))/2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "so |